Let the largest three numbers be $a,b,c$ in decreasing order of value, and let the smallest three numbers be $x,y,z$ in ascending order of value. By the condition, $a+b+c \le a$, thus $b+c \le 0$. Then $b \le -c$. But as stated in the above condition, $a \ge b \ge c$, thus $-c \ge c$ so $c \le 0$. So there are at most two positive numbers. Also by the same condition, $x+y+z \ge x$, thus $y+z \ge 0$. Then $y \ge -z$. But $x \le y \le z$, thus $z \ge -z$ so $z \ge 0$. So there are at most two negative numbers, which means there are at least 2007 zeroes. However, this means that $c = z = 0$, since $c$ is the third largest number and $z$ is the third smallest number. Thus $b + 0 \le 0$ and $y + 0 \ge 0$, so $b \le 0$ and $y \ge 0$. By the same argument, we can show that $b = y = 0$, so this only leaves $a$ and $x$ to be the non-zero numbers, which means there are at least $\boxed{2009}$ zeroes. This can be achieved with the numbers $t$, $-t$, and 2009 zeroes, for any number $t$.